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Let $I=[a,b]$ and let $f:I\to {\mathbb R}$ be a (not necessarily continuous) function with the property that for every $x∈I$, the function $f$ is bounded on a neighborhood $V_{d_x}(x)$ of $x$. Prove that $f$ is bounded on $I$.

Thus far I have that,

For all $n∈I$ there exist $x_n∈[a,b]$ such that $|f(x_n)|>n$. By the Bolzano Weierstrass theorem since $I$ is bounded we have the sequence $X=(x_n)$ is bounded. This implies there is a convergent sub-sequence $X'=(x_{n_r})$ of $X$ that converges to $c$, $c∈[a,b]$. Since $I$ is closed and the element of $X'$ belongs to $I$, it follows from a previous theorem that I proved that $c∈I$. Here is where I get stuck, I want to use that the function $f$ is bounded on a neighborhood $V_{d_x}(x)$ somehow to show that $f$ is bounded on $I$. I'm not sure how to proceed.

$f$ is bounded on $I$ means if there exist a d-neighborhood $V_d(c)$ of $c$ and a constant $M>0$ such that we have $|f(x)|\leq M$ for all $x$ in $A ∩ V_d(c)$.

I would like to do try a proof by contradiction somehow.

Quaternary
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  • You put $x_{n_r}$ back to $xn_r$. Could you explain what your $xn_r$ mean? –  Mar 22 '12 at 19:03
  • It was a mistake on my part when I edited it originally. x_nr is a subsequence. Please change it back to the way you edited it. Sorry! I didn't notice my edit changed that. – Quaternary Mar 22 '12 at 19:04
  • I don't understand what you mean by V_d_x(x). Do you want $V_{d_x}(x)$? –  Mar 22 '12 at 19:05
  • Yes that is precisely what I meant. Thanks again! – Quaternary Mar 22 '12 at 19:06
  • You should have done the $\LaTeX$ stuff yourself. It will be more readable. –  Mar 22 '12 at 19:10
  • I apologize. I'm not familiar with latex. I'll teach it to myself for future purposes. – Quaternary Mar 22 '12 at 19:10
  • Thanks so much guys, I am much more enlightened about this problem with your help. I would like to accept all of your answers! – Quaternary Mar 22 '12 at 19:28

5 Answers5

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This is a simple application of the Heine-Borel theorem. For each $x \in I$ there is a neighbourhood $V_x$ of $x$ and a constant $M_x$ such that $|f(y)| \leq M_x$ for $y \in V_x$. The sets $V_x$ cover $I$ so, by the Heine-Borel theorem, there is a finite subcover $I = V_{x_1} \cup \ldots \cup V_{x_n}$ with $x_i \in I$. Then $f$ is bounded on $I$ by the maximum of $M_{x_1},\ldots,M_{x_n}$.

EDIT: You can also do it by contradiction, without using the Heine-Borel theorem. If $f$ is unbounded then there is a sequence $(x_n)$ such that $\{|f(x_i)| : i\in \mathbb N\}$ is unbounded. Since $I$ is bounded and closed, by the Bolzano-Weierstrass theorem, one can assume that $(x_n)$ is convergent to some $x \in I$. There is a neighbourhood $V_x$ of $x$ such that $f$ is bounded on $V_x$. By the definition of convergence, all but finitely many elements of $(x_n)$ are in $V_x$. But then $\{ |f(x_i)| : i \in \mathbb N \}$ is bounded which is a contradiction.

marlu
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  • Unfortunately, our class has not proven the Heine-Borel Theorem so I cannot use that result. – Quaternary Mar 22 '12 at 19:03
  • here, an unbounded sequence can have a convergent subsequence too , so i dont think we have a contradiction – Joseph Rock Jun 10 '20 at 00:34
  • If I understand correctly, you are concerned about the correctness of the proof because we didn't use the boundedness of $I$. But we did! It is needed to apply the Bolzano-Weierstrass theorem. Without the boundedness, the sequence $(x_i)$ in $I$ could have a convergent subsequence, as you say, but it wouldn't be guaranteed – marlu Jun 11 '20 at 01:20
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Suppose that $f$ is unbounded. Then there exists $x_n \in [a,b]$ with (for example) $\lim_{n \to \infty}f(x_n)=\infty$. Because the sequence $x_n$ is bounded, it has a convergent subsequence, denote it also $x_n \to x_0 \in [a,b]$. Use the fact that $f$ is bounded near $x_0$ to derive a contradiction.

It is just what you did, but you denoted the limit with $c$. The idea is to take a neighborhood of this point, and notice that in any neighborhood there are almost all terms of the sequence $x_n$, for which the values of $f$ are very large.

Beni Bogosel
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If you cannot use the Heine-Borel theorem, argue via sup. Here's a sketch:

Let $A= \{ x \in I : f \text{ is bounded in } [a,x] \}$. Then $A$ is not empty because $a\in A$. Also, $A$ is bounded above because $A\le b$.

Prove that if $x\in A$ and $x<b$ then $x+h\in A$ for some $h>0$ using that $f$ is locally bounded at $x$. This means that no $x< b$ is an upper bound for $A$, which implies that $b=\sup A$.

Finally, using that $f$ is locally bounded at $b$, argue that $b \in A$, thus proving that $f$ is bounded in $I$.

This proof appears in Spivak's Calculus.

lhf
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Proof by contradiction:

if we suppose that $f$ is unbounded in a neighbourhood $V_{x_0}$ of $x_0\in I$ then there exists a sequence $(Z_n)\in V_{x_0}$, $Z_n\to x_0$ and $|f(Z_n)|\to +\infty$ and that is contradictory with the fact that $f$ is locally bounded over $I$.

Abdelmajid Khadari
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So, by hypothesis, you have an open ball $V_d(c)$ and $M$ such that $|f(x)|\le M$ for all $x\in I\cap V_d(c)$ (I think you meant $I$ here, not $V$). But you know that the sequence $(x_{n_r})$ converges to $c$. This means that there is an $m$ so that $x_{n_r}\in I\cap V_d(c)$ whenever $r>m$. Now from the way you selected the $x_n$, what can you conclude?

David Mitra
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